Hi Ishika

It is quite easy dear .

We have, Resistance = R , given in the problem

Now the charge starts from 0 and goes to positive peak at t= a/2b, after that it starts decreasing and crosses zero at t= a/b and then goes towards the negative infinity.
Then,heat produced as a function of t will be -

Q = at - bt

I = dQ/dt = a - 2bt

At t = 0 Q = 0 I = 0

Total heat produced in resistance R

H = IRdt = R(a-2bt)dt
= (a+4bt-4abt) dt
=R[at+4bt/3 -4abt/2]
=R[aa/2b+4b/3a/8b -4ab/2a/4b]
= aR/b[ 1/2+1/6-1/2]
= aR/6b

Thus, the total heat produced in R is aR/6b

I hope that you find this helpful

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